Question: You have found the following ages (in years) of all 6 lions at your local zoo: $ 7,\enspace 4,\enspace 11,\enspace 4,\enspace 10,\enspace 18$ What is the average age of the lions at your zoo? What is the standard deviation? You may round your answers to the nearest tenth.
Solution: Because we have data for all 6 lions at the zoo, we are able to calculate the population mean $({\mu})$ and population standard deviation $({\sigma})$ To find the population mean , add up the values of all $6$ ages and divide by $6$ $ {\mu} = \dfrac{\sum\limits_{i=1}^{{N}} x_i}{{N}} = \dfrac{\sum\limits_{i=1}^{{6}} x_i}{{6}} $ $ {\mu} = \dfrac{7 + 4 + 11 + 4 + 10 + 18}{{6}} = {9\text{ years old}} $ Find the squared deviations from the mean for each lion. Age $x_i$ Distance from the mean $(x_i - {\mu})$ $(x_i - {\mu})^2$ $7$ years $-2$ years $4$ years $^2$ $4$ years $-5$ years $25$ years $^2$ $11$ years $2$ years $4$ years $^2$ $4$ years $-5$ years $25$ years $^2$ $10$ years $1$ year $1$ year $^2$ $18$ years $9$ years $81$ years $^2$ Because we used the population mean $({\mu})$ to compute the squared deviations from the mean , we can find the variance $({\sigma^2})$ , without introducing any bias, by simply averaging the squared deviations from the mean $ {\sigma^2} = \dfrac{\sum\limits_{i=1}^{{N}} (x_i - {\mu})^2}{{N}} $ $ {\sigma^2} = \dfrac{{4} + {25} + {4} + {25} + {1} + {81}} {{6}} $ $ {\sigma^2} = \dfrac{{140}}{{6}} = {23.33\text{ years}^2} $ As you might guess from the notation, the population standard deviation $({\sigma})$ is found by taking the square root of the population variance $({\sigma^2})$ ${\sigma} = \sqrt{{\sigma^2}}$ $ {\sigma} = \sqrt{{23.33\text{ years}^2}} = {4.8\text{ years}} $ The average lion at the zoo is 9 years old. There is a standard deviation of 4.8 years.